Eigenvalue of Real Vector Space Proof

I'm trying to understand a proof from my textbook. They are showing that every operator on a finite dimensional real vector space with odd dimension has an eigenvalue. I already know that every operator on a finite dimensional real vector space has an invariant subspace of dimension two or one...

We have , which is the finite dimensional vector space, and , which is an operator on V. There is a subspace of that is invariant and has dimension 2 (otherwise we are done). There is some other subspace such that .

Anway, here is what I don't understand. Don't we know that either is invariant under or is non-injective (and has eigenvalue 0?). If and , then since U is invariant under T (and T is injective or else 0 is an eigenvalue) there is a such that , which shows that isn't injective?

The short version: if W is not invariant under T, then must T have eigenvalue 0?
Thanks.

I'm trying to understand a proof from my textbook. They are showing that every operator on a finite dimensional real vector space with odd dimension has an eigenvalue. I already know that every operator on a finite dimensional real vector space has an invariant subspace of dimension two or one...

We have , which is the finite dimensional vector space, and , which is an operator on V. There is a subspace of that is invariant and has dimension 2 (otherwise we are done). There is some other subspace such that .

Anway, here is what I don't understand. Don't we know that either is invariant under or is non-injective (and has eigenvalue 0?). If and , then since U is invariant under T (and T is injective or else 0 is an eigenvalue) there is a such that , which shows that isn't injective?

The short version: if W is not invariant under T, then must T have eigenvalue 0?
Thanks.

I think this is easier than that (if this isn't satisfactory, and I'm being a jerk for not reading your post and answering your actual question...feel free to let me know...and I'll read it more carefully :P)

See if you can see why the following lemma finishes the problem

Theorem: Let be a real polynomial and , then has a real zero.Proof: Let

Since if and only if we may assume without loss of generality that is monic. Note then that and . Thus, there exists some such that and . Thus, and . Thus, by the Intermediate Value Theorem there exists some such that . The conclusion follows.

I had typed a really long message here... but I deleted it. Here is the abridged version (I tend to type too much )

Because of the approach of my course and textbook, I have no definition of a characteristic polynomial or determinant of an operator on a real vector space. I tell you this because I suspect that I need these concepts to use your theorem... if not, let me know, and I'll keep trying.

Thanks again...

Last edited by billa; December 8th 2010 at 05:57 PM.
Reason: I found it.